j2+48=77

Solution for j2+48=77 equation:

j2+48=77

We move all terms to the left:

j2+48-(77)=0

We add all the numbers together, and all the variables

j^2-29=0

a = 1; b = 0; c = -29;
Δ = b2-4ac
Δ = 02-4·1·(-29)
Δ = 116
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{116}=\sqrt{4*29}=\sqrt{4}*\sqrt{29}=2\sqrt{29}$
$j_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{29}}{2*1}=\frac{0-2\sqrt{29}}{2}
=-\frac{2\sqrt{29}}{2}
=-\sqrt{29}
$
$j_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{29}}{2*1}=\frac{0+2\sqrt{29}}{2}
=\frac{2\sqrt{29}}{2}
=\sqrt{29}
$